[Math] Proof Disjunctive Syllogism using Natural Deduction

logicnatural-deduction

So, how can I have to prove this using natural deduction:
$\lnot p, p \lor q \vdash q$

What I did is:

  1. $\lnot p$
  2. $p \lor q$
  3. p assumption
  4. $\bot$ from 1&3
  5. q from 4

Is it ok 100% ? What can I do to make it perfect ?
Thanks!

Best Answer

No, it is not.

You have a disjunction as 2nd premise : thus you have to consider both disjuncts with $(\lor \text E)$.

The first sub-case, with $p$ as assumption, is Ok.

You have to add the second sub-case, with $q$ as assumption, in which case the conclusion $q$ is immediate.

Then, having derived $q$ in both cases, you can use $(\lor \text E)$ and conclude.

The flaw in your derivation is that you have the undischarged assumption 3. Thus, what your derivation amounts to is really :

$¬p, p∨q, p \vdash q$.

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